Determining and justifying the validity of an argument












2












$begingroup$


Context: Question made up by uni lecturer



Original statement: There exists two positive real numbers $x$ and $y$ such that for all positive integers $z$, $frac{x}{y}>z$.



So the question was to find the negation of the statement, and then determine whether the original statement or its negation was true.



I found its negation to be: For all positive real numbers $x$ and $y$, there exists a positive integer $z$ such that $frac{x}{y}le z$.



The lecturer's solution to the question says that the negation is true since for any positive reals $x$ and $y$, you can choose $z$ to equal the ceiling of $frac{x}{y}$.



When I attempted the question myself, I said that the original statement is true because you can take $x=z+1$ (which would be a positive integer that still belongs to the set of all positive real numbers) and $y=1$ (which is a positive real number), as $frac{x}{y}=frac{z+1}{1}=z+1>z$.



Can someone please help me to see the error in my answer.



Thanks






discrete-mathematics







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Context: Question made up by uni lecturer



    Original statement: There exists two positive real numbers $x$ and $y$ such that for all positive integers $z$, $frac{x}{y}>z$.



    So the question was to find the negation of the statement, and then determine whether the original statement or its negation was true.



    I found its negation to be: For all positive real numbers $x$ and $y$, there exists a positive integer $z$ such that $frac{x}{y}le z$.



    The lecturer's solution to the question says that the negation is true since for any positive reals $x$ and $y$, you can choose $z$ to equal the ceiling of $frac{x}{y}$.



    When I attempted the question myself, I said that the original statement is true because you can take $x=z+1$ (which would be a positive integer that still belongs to the set of all positive real numbers) and $y=1$ (which is a positive real number), as $frac{x}{y}=frac{z+1}{1}=z+1>z$.



    Can someone please help me to see the error in my answer.



    Thanks






    discrete-mathematics







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Context: Question made up by uni lecturer



      Original statement: There exists two positive real numbers $x$ and $y$ such that for all positive integers $z$, $frac{x}{y}>z$.



      So the question was to find the negation of the statement, and then determine whether the original statement or its negation was true.



      I found its negation to be: For all positive real numbers $x$ and $y$, there exists a positive integer $z$ such that $frac{x}{y}le z$.



      The lecturer's solution to the question says that the negation is true since for any positive reals $x$ and $y$, you can choose $z$ to equal the ceiling of $frac{x}{y}$.



      When I attempted the question myself, I said that the original statement is true because you can take $x=z+1$ (which would be a positive integer that still belongs to the set of all positive real numbers) and $y=1$ (which is a positive real number), as $frac{x}{y}=frac{z+1}{1}=z+1>z$.



      Can someone please help me to see the error in my answer.



      Thanks






      discrete-mathematics







      share|cite|improve this question









      $endgroup$




      Context: Question made up by uni lecturer



      Original statement: There exists two positive real numbers $x$ and $y$ such that for all positive integers $z$, $frac{x}{y}>z$.



      So the question was to find the negation of the statement, and then determine whether the original statement or its negation was true.



      I found its negation to be: For all positive real numbers $x$ and $y$, there exists a positive integer $z$ such that $frac{x}{y}le z$.



      The lecturer's solution to the question says that the negation is true since for any positive reals $x$ and $y$, you can choose $z$ to equal the ceiling of $frac{x}{y}$.



      When I attempted the question myself, I said that the original statement is true because you can take $x=z+1$ (which would be a positive integer that still belongs to the set of all positive real numbers) and $y=1$ (which is a positive real number), as $frac{x}{y}=frac{z+1}{1}=z+1>z$.



      Can someone please help me to see the error in my answer.



      Thanks






      discrete-mathematics




      discrete-mathematics






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Mar 26 at 13:14









      Ruby PaRuby Pa

      376




      376






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          The problem with your reasoning is the order. There exist $x,y$ positive integers such that for all positive integers $z$ we have $frac{x}{y}>z$, so first you must pick an $x$ and a $y$, and then you must test whether $frac{x}{y}>z$ for every positive integer $z$. Therefore you can't define $x=z+1$ as when you pick $x$ you don't know $z$ yet.






          share|cite|improve this answer









          $endgroup$





















            3












            $begingroup$


            "There exists two positive real numbers $x$ and $y$ such that for all positive integer $z$ : $dfrac x y > z$"




            The intuition says that it is False. $dfrac x y$ is a positive real; thus, the statement amounts to asserting that there is a real that is greater than every integer, which is not.



            You reasoning is wrong because you have swapped the choice of the numbers : you start from $z$ and choose $x$ and $y$ accordingly.



            The negation of the original statement is : $forall x forall y exists z (dfrac y y le z)$.



            Thus, choose $x$ and $y$ positive whatever and what you get is a new positive real $dfrac x y$.



            Now, you have to choose an integer $z$ (obviously positive) that is greater-or-equal to $dfrac x y$.



            And this must be always possible, because $dfrac x y$ is a number $r.r_1 r_2 r_3 ldots$.



            Consider as $z$ the number $r+1$ and it's done.






            share|cite|improve this answer











            $endgroup$














              Your Answer





              StackExchange.ifUsing("editor", function () {
              return StackExchange.using("mathjaxEditing", function () {
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              });
              });
              }, "mathjax-editing");

              StackExchange.ready(function() {
              var channelOptions = {
              tags: "".split(" "),
              id: "69"
              };
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function() {
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled) {
              StackExchange.using("snippets", function() {
              createEditor();
              });
              }
              else {
              createEditor();
              }
              });

              function createEditor() {
              StackExchange.prepareEditor({
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              });


              }
              });














              draft saved

              draft discarded


















              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3163151%2fdetermining-and-justifying-the-validity-of-an-argument%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3












              $begingroup$

              The problem with your reasoning is the order. There exist $x,y$ positive integers such that for all positive integers $z$ we have $frac{x}{y}>z$, so first you must pick an $x$ and a $y$, and then you must test whether $frac{x}{y}>z$ for every positive integer $z$. Therefore you can't define $x=z+1$ as when you pick $x$ you don't know $z$ yet.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                The problem with your reasoning is the order. There exist $x,y$ positive integers such that for all positive integers $z$ we have $frac{x}{y}>z$, so first you must pick an $x$ and a $y$, and then you must test whether $frac{x}{y}>z$ for every positive integer $z$. Therefore you can't define $x=z+1$ as when you pick $x$ you don't know $z$ yet.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  The problem with your reasoning is the order. There exist $x,y$ positive integers such that for all positive integers $z$ we have $frac{x}{y}>z$, so first you must pick an $x$ and a $y$, and then you must test whether $frac{x}{y}>z$ for every positive integer $z$. Therefore you can't define $x=z+1$ as when you pick $x$ you don't know $z$ yet.






                  share|cite|improve this answer









                  $endgroup$



                  The problem with your reasoning is the order. There exist $x,y$ positive integers such that for all positive integers $z$ we have $frac{x}{y}>z$, so first you must pick an $x$ and a $y$, and then you must test whether $frac{x}{y}>z$ for every positive integer $z$. Therefore you can't define $x=z+1$ as when you pick $x$ you don't know $z$ yet.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 26 at 13:33









                  Floris ClaassensFloris Claassens

                  1,33229




                  1,33229























                      3












                      $begingroup$


                      "There exists two positive real numbers $x$ and $y$ such that for all positive integer $z$ : $dfrac x y > z$"




                      The intuition says that it is False. $dfrac x y$ is a positive real; thus, the statement amounts to asserting that there is a real that is greater than every integer, which is not.



                      You reasoning is wrong because you have swapped the choice of the numbers : you start from $z$ and choose $x$ and $y$ accordingly.



                      The negation of the original statement is : $forall x forall y exists z (dfrac y y le z)$.



                      Thus, choose $x$ and $y$ positive whatever and what you get is a new positive real $dfrac x y$.



                      Now, you have to choose an integer $z$ (obviously positive) that is greater-or-equal to $dfrac x y$.



                      And this must be always possible, because $dfrac x y$ is a number $r.r_1 r_2 r_3 ldots$.



                      Consider as $z$ the number $r+1$ and it's done.






                      share|cite|improve this answer











                      $endgroup$


















                        3












                        $begingroup$


                        "There exists two positive real numbers $x$ and $y$ such that for all positive integer $z$ : $dfrac x y > z$"




                        The intuition says that it is False. $dfrac x y$ is a positive real; thus, the statement amounts to asserting that there is a real that is greater than every integer, which is not.



                        You reasoning is wrong because you have swapped the choice of the numbers : you start from $z$ and choose $x$ and $y$ accordingly.



                        The negation of the original statement is : $forall x forall y exists z (dfrac y y le z)$.



                        Thus, choose $x$ and $y$ positive whatever and what you get is a new positive real $dfrac x y$.



                        Now, you have to choose an integer $z$ (obviously positive) that is greater-or-equal to $dfrac x y$.



                        And this must be always possible, because $dfrac x y$ is a number $r.r_1 r_2 r_3 ldots$.



                        Consider as $z$ the number $r+1$ and it's done.






                        share|cite|improve this answer











                        $endgroup$
















                          3












                          3








                          3





                          $begingroup$


                          "There exists two positive real numbers $x$ and $y$ such that for all positive integer $z$ : $dfrac x y > z$"




                          The intuition says that it is False. $dfrac x y$ is a positive real; thus, the statement amounts to asserting that there is a real that is greater than every integer, which is not.



                          You reasoning is wrong because you have swapped the choice of the numbers : you start from $z$ and choose $x$ and $y$ accordingly.



                          The negation of the original statement is : $forall x forall y exists z (dfrac y y le z)$.



                          Thus, choose $x$ and $y$ positive whatever and what you get is a new positive real $dfrac x y$.



                          Now, you have to choose an integer $z$ (obviously positive) that is greater-or-equal to $dfrac x y$.



                          And this must be always possible, because $dfrac x y$ is a number $r.r_1 r_2 r_3 ldots$.



                          Consider as $z$ the number $r+1$ and it's done.






                          share|cite|improve this answer











                          $endgroup$




                          "There exists two positive real numbers $x$ and $y$ such that for all positive integer $z$ : $dfrac x y > z$"




                          The intuition says that it is False. $dfrac x y$ is a positive real; thus, the statement amounts to asserting that there is a real that is greater than every integer, which is not.



                          You reasoning is wrong because you have swapped the choice of the numbers : you start from $z$ and choose $x$ and $y$ accordingly.



                          The negation of the original statement is : $forall x forall y exists z (dfrac y y le z)$.



                          Thus, choose $x$ and $y$ positive whatever and what you get is a new positive real $dfrac x y$.



                          Now, you have to choose an integer $z$ (obviously positive) that is greater-or-equal to $dfrac x y$.



                          And this must be always possible, because $dfrac x y$ is a number $r.r_1 r_2 r_3 ldots$.



                          Consider as $z$ the number $r+1$ and it's done.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Mar 26 at 13:44

























                          answered Mar 26 at 13:29









                          Mauro ALLEGRANZAMauro ALLEGRANZA

                          67.7k449117




                          67.7k449117






























                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to Mathematics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid …



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3163151%2fdetermining-and-justifying-the-validity-of-an-argument%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              -

                              Popular posts from this blog

                              How to change which sound is reproduced for terminal bell?

                              Image
                              3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
                              Read more

                              Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

                              Image
                              up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
                              Read more

                              Can I use Tabulator js library in my java Spring + Thymeleaf project?

                              Image
                              0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
                              Read more